document.write( "Question 498563: Laura Dawson can get to school in 1/4 hour if she rides her bike. It takes her 3/4 if she walks. Her speed when walking is 6 miles per hour slower than her speed when riding.What is her speed when riding? \n" ); document.write( "
Algebra.Com's Answer #337067 by stanbon(75887)\"\" \"About 
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Laura Dawson can get to school in 1/4 hour if she rides her bike. \r
\n" ); document.write( "\n" ); document.write( "It takes her 3/4 hr. if she walks. \r
\n" ); document.write( "\n" ); document.write( "Her speed when walking is 6 miles per hour slower than her speed when riding.
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\n" ); document.write( "Walking DATA:
\n" ); document.write( "time = 3/4 hr ; speed = r mph ; distance = (3/4)r miles
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\n" ); document.write( "Riding DATA:
\n" ); document.write( "time = 1/4 hr ; speed = r+6 mph ; distance = (1/4)(r+6)
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\n" ); document.write( "Equation:
\n" ); document.write( "distance = distance
\n" ); document.write( "(3/4)r = (1/4)(r+6)
\n" ); document.write( "3r = r+6
\n" ); document.write( "2r = 6
\n" ); document.write( "r = 3 mph (walking speed)\r
\n" ); document.write( "\n" ); document.write( "What is her speed when riding?
\n" ); document.write( "r+6 = 9 mph (riding speed)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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